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Ivanshal [37]
3 years ago
15

A rubber ball weighs 49 N.. a) What is the mass of the ball? (Answer: 5.0 kg). b) What is the acceleration of the ball if an upw

ard force of 69 N is applied? (Answer: 4.0 m/s/s). For a) I used the equation W = Fg = mg, which becomes 49 = m(9.8). This approximates m to 5.0 kg.
---
For b) I used the equation
ΣFy=Fa−W=ma
which becomes 69-49 = 20 N.
I then use equation F = ma which becomes 20 = 5.0a, which makes a
4.0m/s2
Physics
1 answer:
weqwewe [10]3 years ago
4 0
The answer you presented is correct.

For a) I used the equation W = Fg = mg, which becomes 49 = m(9.8). This approximates m to 5.0 kg.
<span>For b) I used the equation
</span><span>ΣFy=Fa−W=ma</span> which becomes 69-49 = 20 N.
I then use equation F = ma which becomes 20 = 5.0a, which makes a <span>4.0m/<span>s2</span></span>
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The normal force is the supporting force that is exerted on an object that is in contact with another stable object.

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3 years ago
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What’s the voltage of a battery in a circuit with resistance of 3 ohms and current of 5 amps?
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3 years ago
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3 years ago
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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
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